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Guest Message by DevFuse


Member Since --
Offline Last Active Today, 05:48 AM

#338863 Max. # of convex/concave hexagons possible using 6 points

Posted by bonanova on 01 July 2014 - 07:43 AM

Spoiler for a few more


Spoiler for Agree with your result

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#338647 Four numbers equals two dozen

Posted by bonanova on 18 June 2014 - 06:58 AM

Spoiler for more shenanigans

Cute. Honorable mention.
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#338646 Four numbers equals two dozen

Posted by bonanova on 18 June 2014 - 06:56 AM

Spoiler for Possibility

Creative. But sqrt is not available. :(
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#338645 Four numbers equals two dozen

Posted by bonanova on 18 June 2014 - 06:38 AM


Board Guideline 6 suggests being positive. If you find a puzzle not to your liking, try another. Authors are free to compose flavor text to suit their own particular style. Your kid sister can type the word Google. Most puzzle books have an answer section. Not really the point of buying the book, tho, is it? Please consider the reason you joined this site. And do read the posting guidelines. Thanks.
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#338588 Pendulum Watch

Posted by bonanova on 07 June 2014 - 08:21 AM

Spoiler for Looks like

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#338563 Paint Color Picker

Posted by bonanova on 04 June 2014 - 04:01 PM

Well, you named this "paint color."

Paint is formed of pigment, where CMYK is relevant.

In the RGB space, I didn't find any of them remarkable.


Spoiler for Except that

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#338527 DENSA entrance exam question

Posted by bonanova on 02 June 2014 - 04:31 AM

Until this posting, it was a fairly well-kept secret that I am a card-carrying member of the DENSA society. That is not a typo. DENSA is an acronym for Diversely Educated, Not Seriously Affected.
DENSA membership requires supplying the correct answer to the following multiple-choice question.
1. How many stars are there?
      a. All of the above.
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#338421 The "aha!" problems 8: Reflect on this

Posted by bonanova on 22 May 2014 - 12:22 PM

A ray of light encounters a pair of angled mirrors from the left, as shown, a distance d from one mirror and at an angle a from the other mirror. If angle a is exactly 22.5o, how close will the ray get to their intersection at point O before it eventually exits again to the left? The drawing may not be to scale.


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#338401 shortest distance

Posted by bonanova on 21 May 2014 - 06:59 AM

Spoiler for For starters

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#338228 Adding really simple fractions

Posted by bonanova on 02 May 2014 - 11:34 AM

Johnny was asked to write the sum of the two fractions p/n and q/n where p, q and n are single-digit numbers.

He remembered that the numerators were to be added, but he thought maybe the denominators should be added as well.

Also, he decided that putting digits next to each other could suggest addition.

So his first try was to say that


p/n + q/n = pq/nn. (where pq and nn are two-digit numbers.)


Then he remembered the result should not depend on the order of addition,

but he saw that p/n + q/n = qp/nn was just as bad. So he decided to use them both.

His final representation of the sum was


p/n + q/n = (pq + qp)/nn.


The puzzle asks: Let p, q and n be chosen at random from {1, 2, 3, 4, 5, 6, 7, 8, 9}.

What is the expected error in Johnny's formula?

  • -1

#338150 The "aha!" problems 5. Four-point square

Posted by bonanova on 28 April 2014 - 08:05 AM

Take any four points A, B, C, D in the plane, no three of which are collinear.

They describe a unique quadrilateral, if we take the points as being its vertices.


But they also describe a square, if we require only that the points lie on its sides.

Using a compass and straightedge, construct a square such that the four points lie, one each, on its sides

Edit: or the extensions of its sides.

. . . . . . . . . . . . . . . 4point_square_a.jpg

Hint: The points are not special.

Draw four similar points and do the construction on a sheet of paper if that helps.

The answer would then be to describe the process.

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#338131 The "aha!" problems - 3 minimum areas

Posted by bonanova on 27 April 2014 - 07:27 AM

A while back I started a series of puzzles that seem difficult but

have not-so-difficult answers. I can't find any of them, so I don't

know how many exist. There must have been at least two, so I'll

number this one 3. Have fun.


A monotonic increasing function f (x) is cut above and below by horizontal lines that intersect it at  f (x1) and f (x2).

A vertical line is drawn through a point on the curve (black dot) between x1 and x2.

The curve and the lines define the green and red areas shown on the figure.


You are to find the point on the curve that minimizes the sum of these areas.


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#337991 More triangles in circles

Posted by bonanova on 04 April 2014 - 03:28 PM

Edit: It's actually more complicated than that hideous monstrosity... I left out that you would need to normalize the probabilities to sum to 1 instead of using raw squares of distances from the center.


You'd have the integrals you describe in the numerator of a fraction, divided  by the same integrals without the HasOrigin part in the denominator.

I think that would do the normalization. Ugh, agreed.

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#337815 More triangles in circles

Posted by bonanova on 16 March 2014 - 06:13 AM

While our collective brain trust ponders the nature of triangles defined

by three uniformly random points chosen inside a circle, specifically

the mode of their areas, we ask another question.


Recalling that the mean and median areal coverage of its circle's area are

about 7.4% and 5.4%, what is the probability that a random triangle covers

its circle's center?

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#337814 More random numbers

Posted by bonanova on 16 March 2014 - 05:42 AM

Inspired by BMAD's puzzle, consider choosing four random numbers on [0,1].

If you sort them lowest to highest, what is the expected value of the third number?

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